Polynomial Space [HMU06,Chp.11b] • The classes PS and NPS • Relationship to Other Classes • Equivalence PS = NPS • Quantified Boolean Formulas • A PS-Complete Problem 1
Polynomial Space Machines & Classes 2
Polynomial-Space-Bounded TM’s A TM M is said to be polyspace- bounded if there is a polynomial p(n) such that, given input of length n, M never uses more than p(n) cells of its tape. L(M) is in the class polynomial space , or PS , also called PSPACE. 3
NonDeterministic Polyspace If we allow a TM M to be nondeterministic but to use only p(n) tape cells in any sequence of ID’s when given input of length n, we say M is a nondeterministic polyspace-bounded TM. And L(M) is in the class nondeterministic polyspace , or NPS . 4
Relationship to Other Classes Obviously, P ⊆ PS and NP ⊆ NPS . If you use polynomial time, you cannot reach more than a polynomial number of tape cells. Alas, it is not even known whether P = PS or NP = PS . On the other hand, we shall show PS = NPS . [cf. believed P≠NP] 5
Exponential Polytime Classes A DTM M runs in exponential polytime if it makes at most c p(n) steps on input of length n, for some constant c and some polynomial p. Say L(M) is in the class EXP . If M is an NTM instead, say L(M) is in the class NEXP ( nondeterministic exponential polytime ). 6
More Class Relationships P ⊆ NP ⊆ PS ⊆ EXP at least one of these is proper. a diagonalization proof shows that P ≠ EXP . PS ⊆ EXP requires proof. Key Point: A polyspace-bounded TM has only c p(n) different ID’s. We can count to c p(n) in polyspace and stop it after it surely repeated an ID. 7
Proof PS ⊆ EXP Let M be a p(n)-space bounded DTM with s states and t tape symbols. Assume M has only one semi-infinite tape. The number of possible ID’s of M is s ⨯ p(n) ⨯ t p(n) . Tape Positions of contents States tape head 8
Proof PS ⊆ EXP – (2) Note that (t+ 1) p(n)+ 1 > p(n)t p(n) . Use binomial expansion (t+ 1) p(n)+ 1 = t p(n)+ 1 + (p(n)+ 1)t p(n) + … Also, s = (t+ 1) c , where c = log t+ 1 s. Thus, s ⨯ p(n) ⨯ t p(n) < (t+ 1) p(n)+ 1+ c . We can count to the maximum number of ID’s on a separate tape using base t+ 1 and p(n)+ 1+ c cells – a polynomial. 9
Proof PS ⊆ EXP – (3) Redesign M to have a second tape and to count on that tape to s ⨯ p(n) ⨯ t p(n) . The new TM M’ is polyspace bounded. M’ halts if its counter exceeds s ⨯ p(n) ⨯ t p(n) . If M accepts, it does so without repeating an ID. Thus, M’ is exponential-polytime bounded, proving L(M) is in EXP . 10
Equivalence PS = NPS 11
Savitch’s Theorem: PS = NPS Key Idea: a polyspace NTM has “only” c p(n) different ID’s it can enter. Implement a deterministic, recursive function that decides, about the NTM, whether I ⊦ * J in at most m moves. Assume m < c p(n) , since if the NTM accepts, it does so without repeating an ID. 12
Savitch’s Theorem – (2) Recursive doubling trick: to tell if I ⊦ * J in < m moves, search for an ID K such that I ⊦ * K and K ⊦ * J, both in < m/2 moves. Complete algorithm: ask if I 0 ⊦ * J in at most c p(n) moves, where I 0 is the initial ID with given input w of length n, and J is any of the ID’s with an accepting state and length < p(n). 13
Recursive Doubling boolean function f(I, J, m) { for (all ID’s K using p(n) tape) if (f(I, K, m/2) && f(K, J, m/2)) return true; return false; } 14
Stack Implementation of f I, J, m I, K, m/2 L, K, m/4 . . . M, N, 1 O(p(n)) O(p(n)) O(p(n)) O(p(n)) space space space space O(p 2 (n)) space 15
Space for Recursive Doubling f(I, J, m) requires space O(p(n)) to store I, J, m, and the current K. m need not be more than c p(n) , so it can be stored in O(p(n)) space. How many calls to f can be active at once? Largest m is c p(n) . 16
Space for Recursive Doubling – (2) Each call with third argument m results in only one call with argument m/2 at any one time. Thus, at most log 2 c p(n) = O(p(n)) calls can be active at any one time. Total space needed by the DTM is therefore O(p 2 (n)) – a polynomial. ∎ 17
PS-Complete Problems A problem P in PS is said to be PS- complete if there is a polytime reduction from every problem in PS to P. Note: it has to be polytime, not polyspace, because: 1. Polyspace can exponentiate the output size. 2. Without polytime, we could not deal with the question P = PS ? 18
What PS-Completeness Buys If some PS-complete problem is: 1. In P , then P = PS . 2. In NP , then NP = PS . 19
Quantified Boolean Formulas 20
Quantified Boolean Formulas We shall meet a PS-complete problem, called QBF : is a given quantified boolean formula true? But first we meet the QBF’s themselves. We shall give a recursive (inductive) definition of QBF’s along with the definition of free/bound variable occurrences. 21
QBF’s – (2) First-order predicate logic, with variables restricted to true/false. Basis: 1. Constants 0 (False) and 1 (True) are QBF’s. 2. A variable is a QBF, and that variable occurrence is free in this QBF. 22
QBF’s – (3) Induction: If E and F are QBF’s, so are: 1. E ∧ F, E ∨ F, and ¬ F. Variables are bound or free as in E or F. 2. ( ∀ x)E and ( ∃ x)E for any variable x. All free occurrences x are bound to this quantifier , and other occurrences of variables are free/bound as in E. Use parentheses to group as needed. Precedence: quantifiers, ¬ , ∧ , ∨ . 23
Example: QBF bound ( ∀ x)( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (x ∧ y) ) bound bound 24
Evaluating QBF’s In general, a QBF is a function from truth assignments for its free variables to { 0, 1} (False/True). Important special case: no free variables; a QBF is either true or false. We shall give the evaluation only for these formulas. 25
Evaluating QBF’s – (2) Induction on the number of operators, including quantifiers. Stage 1: eliminate quantifiers. Stage 2: evaluate variable-free formulas. Basis: 0 operators. Expression can only be 0 or 1, because there are no free variables. Truth value is 0 or 1, respectively. 26
Induction 1. Expression is ¬ E, E ∨ F, or E ∧ F. Evaluate E and F; apply boolean operator to the results. 2. Expression is ( ∀ x)E. Construct E 0 = E with each x bound to this quantifier replaced by 0, and analogously E 1 . E is true iff both E 0 and E 1 are true. 3. Expression is ( ∃ x)E. Same, but E is true iff E 0 or E 1 is true. 27
Example: Evaluation ( ∀ x)( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (x ∧ y) ) Substitute x = 0 for outer quantifier: ( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (0 ∧ y) ) Substitute x = 1 for outer quantifier: ( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (1 ∧ y) ) 28
Example: Evaluation – (2) Let’s follow the x = 0 subproblem: ( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (0 ∧ y) ) Two cases: y = 0 and y = 1. ( ( ∃ x)(x ∨ 0) ) ∧ ¬ (0 ∧ 0) ( ( ∃ x)(x ∨ 1) ) ∧ ¬ (0 ∧ 1) 29
Example: Evaluation – (3) Let’s follow the y = 0 subproblem: ( ( ∃ x)(x ∨ 0) ) ∧ ¬ (0 ∧ 0) Need to evaluate ( ∃ x)(x ∨ 0). x = 0: 0 ∨ 0 = 0. x = 1: 1 ∨ 0 = 1. Hence, value is 1. Answer is 1 ∧ ¬ (0 ∧ 0) = 1. 30
Example: Evaluation – (4) Let’s follow the y = 1 subproblem: ( ( ∃ x)(x ∨ 1) ) ∧ ¬ (0 ∧ 1) Need to evaluate ( ∃ x)(x ∨ 1). x = 0: 0 ∨ 1 = 1. x = 1: 1 ∨ 1 = 1. Hence, value is 1. Answer is 1 ∧ ¬ (0 ∧ 1) = 1. 31
Example: Evaluation – (5) Now we can resolve the (outermost) x = 0 subproblem: ( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (0 ∧ y) ) We found both of its subproblems are true. We only needed one, since the outer quantifier is ∃ y. Hence, 1. 32
Example: Evaluation – (6) Next, we must deal with the x = 1 case: ( ∃ y) ( ( ( ∃ x)(x ∨ y) ) ∧ ¬ (1 ∧ y) ) It also has the value 1, because the subproblem y = 0 evaluates to 1. Hence, the entire QBF has value 1. 33
A PS-Complete Problem: QBF 34
The QBF Problem The problem QBF is: Given a QBF with no free variables, is its value 1 (true)? Theorem: QBF is PS-complete. Comment: What makes QBF extra hard? Alternation of quantifiers. Example: if only ∃ used, then the problem is really SAT. 35
Part I: QBF is in PS Suppose we are given QBF F of length n. F has at most n operators. We can evaluate F using a stack of subexpressions that never has more than n subexpressions, each of length < n. Thus, space used is O(n 2 ). 36
QBF is in PS – (2) Suppose we have subexpression E on top of the stack, and E = G ∨ H. 1. Push G onto the stack. 2. Evaluate it recursively. 3. If true, return true. 4. If false, replace G by H, and return what H returns. 37
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